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NEET · Physics · STD 12 - 2. Electric potential and capacitance

A capacitor of \(2\,\, \mu F\) is charged as shown in the diagram. When the switch \(S\) is turned to position \(2,\) the percentage of its stored energy dissipated is ......\(\%\) 

  1. A \(20\)
  2. B \(75\)
  3. C \(80\)
  4. D \(0\)
Verified Solution

Answer & Solution

Correct Answer

(C) \(80\)

Step-by-step Solution

Detailed explanation

Initially, the energy stored in \(2 \,\mu F\) capacitor is \(U_{i}=\frac{1}{2} C V^{2}=\frac{1}{2}\left(2 \times 10^{-6}\right) V^{2}=V^{2} \times 10^{-6}\, \mathrm{J}\)